what is the external net force exerted on a 3.5 kg papaya which is being pushed across a table and has an acceleration of 2.2 m/s2 to left?
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what is the external net force exerted on a 3.5 kg papaya which is being pushed across a table and has an acceleration of 2.2 m/s2 to left?
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Verified answer
Given: a=2.2m/s^2 to the left
m=3.5kg
Solution: F=ma
=(2.2m/s^2)(3.5kg)
=7.7kg.m/s^2 or 7.7N, to the left